A block slides down a frictionless incline with constant acceleration. After sliding 6.80 m down, it has a speed of 3.80 m/s. Wh
1 answer:
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Missing question: "What is the spring's constant?"
Solution:
The object of mass m=6.89 kg exerts a force on the spring equal to its weight:
When the object is attached to the spring, the displacement of the spring with respect to its equilibrium position is
And by using Hook's law, we can find the constant of the spring:
Solution:
The object of mass m=6.89 kg exerts a force on the spring equal to its weight:
When the object is attached to the spring, the displacement of the spring with respect to its equilibrium position is
And by using Hook's law, we can find the constant of the spring:
Answer:
a)Velocity of car =v=16 m/s
b)Force against the track at point B=1.15*N
Explanation:
Given mass of roller coaster=m=350 kg
Position of A=Ha=25 m
Position of B=Hb=12 m
Net potential energy=mg(ha-hb)
Net potential energy=(350)(9.80)(25-12)
Net potential energy=44590 J
Using energy conservation
net kinetic energy=net potential energy
(1/2)mv^2=mg(ha-hb)
m=350
velocity=v=16 m/s
b)There two force acting,centripetal force upward and gravity downward.
Thus net force acting will be
Net force=(mv^2/r)-mg
Net force=14933.33-3430
Net force=1.15* N